Question in the attachments An object of height 18.4 cm is distance \( s_0 = 89.8 \) cm from a **concave mirror** of focal length 2.13 cm. At \( t = 0 \), the shape of the mirror begins to change, so that the focal length is decreasing at a constant rate:\[\frac{df}{dt} = -3.57 \text{ cm/s}.\]Find the rate at which the height of the image is changing, in cm/s. The sign will indicate if the value of the height of the image is increasing or decreasing.**HINT:** Again, beware of signs!

Using thin lens equation for lens 2 1f=1do+1di1di=1f-1dodi=dofdo-f

An object of height 18.4 cm is distance so = 89.8 cm from a concave mirror of focal length 2.13 cm. Att = 0 the shape of the mirror begins to change, so that the focal length is decreasing at a constant rate: df = -3.57 cm/s. dt Find the rate at which the height of the image is changing, in cm/s. The sign will indicate if the value of the height of the image is increasing or decreasing. HINT: Again, beware of signs!

An object of height 18.4 cm is distance so = 89.8 cm from a concave mirror of focal length 2.13 cm. Att = 0 the shape of the mirror begins to change, so that the focal length is decreasing at a constant rate: df = -3.57 cm/s. dt Find the rate at which the height of the image is changing, in cm/s. The sign will indicate if the value of the height of the image is increasing or decreasing. HINT: Again, beware of signs!